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Find the product of all positive integral values of \(n\) such that \(n^2-35n+306= p\) for some prime number \(p\). Note that there is at least one such \(n\).

 

XD

 Apr 6, 2020
 #1
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This problem seems tricky at first, if you plug in any positive integer, you will always get an even number. There is only one even prime which is 2. When n=16,19, p=2 which are the only values that work. So 16*19= 304 which is your answer.

 

 

Hope it helps!

 Apr 6, 2020

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